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As I understand it, the Landauer limit, $kTln(2)$, is the minimum amount of energy to erase a bit. Is it also the minimum amount to create a bit? I'm asking statistical, like Avogadro's number of bits, i.e., $RTln(2)$.

My thinking: $RTln(2)$ is the work required, probably both ways (create and erase), but I don't know whether that is the same as the energy. By "same" I mean classical like heat, like something that can be used in $E=mc^{2}$.

Ramiro Hum-Sah
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Incredible II
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    How are bits created? Can they be created? – endolith Oct 05 '16 at 02:14
  • @endolith I think the OP will probably agree that he/ she means "initialized" rather than created. It's a well formed question and the OP's intuition, that it does require work to initialize, is correct. – Selene Routley Sep 19 '17 at 13:21

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One doesn't really create bits. One simply changes the state of a physical system to "store" a bit of information. At a microscopic level, all changes are governed by reversible laws, so that the state of the physical storage system before the storage must somehow wind up encoded in the physical state of the environment. This leads to the need to input work to the computer system to keep it at a constant macrostate, as discussed in my answer here. Otherwise put, we need to input work to "forget" the storage system's state before initialization.

So, to initialize (rather than creating) a bit does indeed require the input of a minimum quantity of work of $k\,T\,\log 2$ joules, in accordance with Landauer's principle, so that the storage system and environment's state changes comply with the second law of thermodynamics.

Selene Routley
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To create a bit in some defined state like 0 takes energy $kT\ln(2)+W$, where $W$ is the work required to create whatever the bit is made of (a flip-flop, a particle with spin, etc.). The first factor is just the Landauer cost of setting it in a definite state. As you say for a mole of bits you will have to pay at least $RT\ln(2)$ Joules.

$W$ will be problem dependent, and could in principle be zero: we can just name existing particles with spin as our bits without doing anything to them ("That electron over there will be bit 1, the other one bit 2, and the one over on Mercury bit 3..."). That kind of arbitrary naming still has the problem that we need to remember which electron is which bit, paying a Landauer cost for the naming, but often the indexing of bits can be done implicitly like when they are all spins along a crystal lattice we just found and decided to make our memory.

Since the work $W$ of making a bit typically involves irreversibly changing multiple degrees of freedom while the Landauer limit is about just one degree of freedom, most of the time $W \gg kT\ln(2)$.

Anders Sandberg
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